By Alexander Schmitt

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**Extra info for Algebra I: Commutative Algebra [Lecture notes]**

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N, there is the canonical surjection πk : R −→ R/Ik a −→ [a]k . , [a]n . Note that n Ik . ker(ϕ) = k=1 By the isomorphism theorem, we get an induced injective homomorphism n ϕ : R/ k=1 n Ik −→ R/Ik . k=1 Next, we would like to study when ϕ or, equivalently, ϕ is surjective. , n. , n } : ei ∈ im(ϕ). Let us look at the condition e1 ∈ im(ϕ) in more detail. , n. , n. , n. , n. 8. Operations on Ideals Set n a := n sk = k=2 k=2 (1 − rk ). , n. It follows ϕ(a) = e1 . 11 Lemma. The homomorphism ϕ is surjective if and only if the ideals Ik and Il are coprime, for 1 ≤ k < l ≤ n.

The coefficient of xdeg( fk ) in fk , k ≥ 1. Claim. , ak , ak+1 . , rk ∈ R with k ak+1 = i=1 r i · ai . The definition of the relation “<” implies ∀k : deg( fk ) ≤ deg( fk+1 ). Therefore, we can define k g := fk+1 − i=1 ri · xdeg( fk+1 )−deg( fi ) · fi . 1) But then, deg(g) < deg( fk+1 ). , fk . , fk . , fk , and this is a contradiction. , ak )k≥1 is a non-stationary ascending chain in R. This contradicts the assumption that R is noetherian. 7 Corollary. , a field. , xn] is noetherian, n ≥ 1.

6 Examples. i) We look at the ideals Ix = y, z , Iy = x, z , and Iz = x, y inside k[x, y, z]. Then, V(Ix ) is the x-axis, V(Iy ) the y-axis, and V(Iz ) the z-axis. 8. 2, ii), both V(Ix · Iy · Iz ) and V(Ix ∩ Iy ∩ Iz ) consist of the union of the coordinate axes. ∪ ∪ = Note that Ix · Iy · Iz Ix ∩ Iy ∩ Iz ⊂ I V(Ix ∩ Iy ∩ Iz ) = I V(Ix · Iy · Iz ) . 20, ii). ii) We look at the ideals I1 := y2 − x3 , z and I2 := x, y in the ring k[x, y, z]. 2, iv) inside the (x, y)-plane and V(I2 ) is, as before, the z-axis.